JEE Main 2026 — Mathematics Probability & Statistics

Let $E_1$E1​, $E_2$E2​, and $E_3$E3​ be the events that the candidate travels by bus, scooter, and car, respectively.

The probabilities of choosing these modes of transport are:

$$P(E_1) = \dfrac{2}{5}$$P(E1​)=52​

$$P(E_2) = \dfrac{1}{5}$$P(E2​)=51​

$$P(E_3) = \dfrac{2}{5}$$P(E3​)=52​

Let $A$A be the event that the candidate reaches the examination centre late. The conditional probabilities of reaching late are:

$$P(A|E_1) = \dfrac{1}{5}$$P(A∣E1​)=51​

$$P(A|E_2) = \dfrac{1}{3}$$P(A∣E2​)=31​

$$P(A|E_3) = \dfrac{1}{4}$$P(A∣E3​)=41​

We need to find the probability that the candidate travelled by bus given that they reached late, which is $P(E_1|A)$P(E1​∣A).

Using Bayes' theorem:

$$P(E_1|A) = \dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$P(E1​∣A)=P(E1​)P(A∣E1​)+P(E2​)P(A∣E2​)+P(E3​)P(A∣E3​)P(E1​)P(A∣E1​)​

Substituting the given values:

$$P(E_1|A) = \dfrac{\dfrac{2}{5} \times \dfrac{1}{5}}{\left(\dfrac{2}{5} \times \dfrac{1}{5}\right) + \left(\dfrac{1}{5} \times \dfrac{1}{3}\right) + \left(\dfrac{2}{5} \times \dfrac{1}{4}\right)}$$P(E1​∣A)=(52​×51​)+(51​×31​)+(52​×41​)52​×51​​

$$P(E_1|A) = \dfrac{\dfrac{2}{25}}{\dfrac{2}{25} + \dfrac{1}{15} + \dfrac{1}{10}}$$P(E1​∣A)=252​+151​+101​252​​

Taking the LCM of the denominators $25$25, $15$15, and $10$10, which is $150$150:

$$P(E_1|A) = \dfrac{\dfrac{12}{150}}{\dfrac{12}{150} + \dfrac{10}{150} + \dfrac{15}{150}}$$P(E1​∣A)=15012​+15010​+15015​15012​​

$$P(E_1|A) = \dfrac{12}{12 + 10 + 15} = \dfrac{12}{37}$$P(E1​∣A)=12+10+1512​=3712​

Answer: $\dfrac{12}{37}$3712​